1020. Tree Traversals (25)

1. Tree Traversals (25)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2

#include<stdio.h>#include<stdlib.h>typedef struct TreeNode* BinTree;struct TreeNode{    int data;    BinTree left,right;};BinTree build(int* post,int* in,int n){//有后序遍历和中序遍历，构造二叉树     BinTree root;    if(!n) return NULL;    root=malloc(sizeof(struct TreeNode));    root->data=post[n-1];    int i;    for(i=0;i<n;i++){        if(post[n-1]==in[i]) break;    }    root->left=build(post,in,i);    root->right=build(post+i,in+i+1,n-1-i);    return root;}void bfs(BinTree root,int n,int ans[]){//对二叉树进行宽度优先遍历     int front=0,rear=1;    BinTree q[n];    q[0]=root;    int i=0;    while(front<rear){        BinTree u=q[front++];        ans[i++]=u->data;        if(u->left) q[rear++]=u->left;        if(u->right) q[rear++]=u->right;        }}void remove_tree(BinTree root){     if(root==NULL) return;    remove_tree(root->left);    remove_tree(root->right);    free(root);}int main(){    int N;    scanf("%d",&N);    int post[N],in[N],ans[N];    int i;    for(i=0;i<N;i++){        scanf("%d",&post[i]);    }    for(i=0;i<N;i++){        scanf("%d",&in[i]);    }    BinTree root=build(post,in,N);    bfs(root,N,ans);    remove_tree(root);    for(i=0;i<N;i++){        if(i) printf(" %d",ans[i]);        else printf("%d",ans[i]);    }    return 0; }